The combustion of propane may be described by the chemical equation:
\[ \text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g) \]
How many grams of \(\text{O}_2(g)\) are needed to completely burn 40.4 g of \(\text{C}_3\text{H}_8(g)\)?
Answer (2)
ν C 3 H 8 = μ C 3 H 8 40.4 = = 3 A C + 8 A H 40.4 = 44 40.4 = = 10 9 = 0.9 m o l 1 m o l C 3 H 8 ...5 \ m o l O 2 0.9 m o l C 3 H 8 ... x \ m o l O 2 x = 1 0.9 ∗ 5 = 4.5 m o l O 2 m = μ O 2 ∗ x = 2 ∗ A O ∗ 4.5 = 9 ∗ 8 = 72 g O 2
To completely burn 40.4 g of propane (C₃H₈), approximately 146.56 g of oxygen (O₂) is required. This is calculated by converting grams of propane to moles, using the balanced reaction equation, and converting the moles of O₂ back to grams. Ultimately, the molar ratio indicates that for every mole of propane, 5 moles of oxygen are necessary.
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